Let be a finite field. Then has elements, whereas is the characteristic of and .
Furthermore I have to proof, that for all .

Could anybody of you explain, how to proof the proposition?

Bye,
Lisa

I'll assume p is a prime number. A finite field K of order is the splitting field of over (see here). Since a finite field is perfect, K is a separable extension of . It follows that K is a Galois extension of a field . By the fundamental theorem of Galois theory, , where and Gal(K/F) is a cyclic group of order n generated by a Frobenius automorphism.

Let be a finite field. Then has elements, whereas is the characteristic of and .
Furthermore I have to proof, that for all .

Could anybody of you explain, how to proof the proposition?

Bye,
Lisa

One of the easiest proofs is, perhaps, to note that any field is a vector space over any of its subfields, and then: a finite field of characteristic p is a v.s. over the prime field of char. p, , obviously of finite dimension n, and choosing any basis for this v.s. a simple combinatoric reasoning gives that the number of elements in must be

Take the polynomial . Now you can go with the argument from fields extensions and Galois theory that was given to you by Aliceinwonderland or else use a little group theory: the multiplicative group has order , and we're done.